Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $p = \dfrac{10k + 10}{6k - 12} \times \dfrac{k^2 - 11k + 18}{-5k^2 - 5k} $
Solution: First factor the quadratic. $p = \dfrac{10k + 10}{6k - 12} \times \dfrac{(k - 2)(k - 9)}{-5k^2 - 5k} $ Then factor out any other terms. $p = \dfrac{10(k + 1)}{6(k - 2)} \times \dfrac{(k - 2)(k - 9)}{-5k(k + 1)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 10(k + 1) \times (k - 2)(k - 9) } { 6(k - 2) \times -5k(k + 1) } $ $p = \dfrac{ 10(k + 1)(k - 2)(k - 9)}{ -30k(k - 2)(k + 1)} $ Notice that $(k + 1)$ and $(k - 2)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 10\cancel{(k + 1)}(k - 2)(k - 9)}{ -30k\cancel{(k - 2)}(k + 1)} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $p = \dfrac{ 10\cancel{(k + 1)}\cancel{(k - 2)}(k - 9)}{ -30k\cancel{(k - 2)}\cancel{(k + 1)}} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $p = \dfrac{10(k - 9)}{-30k} $ $p = \dfrac{-(k - 9)}{3k} ; \space k \neq 2 ; \space k \neq -1 $